Prove that for all integers $n$ if $3 \mid n^2$, then $3 \mid n$

Question

Prove that for all integers $n$ if $3$ | $n^2$, then $3$ | $n$.

I figured using contrapostive was the best method by using the definition "an integer $k$ is not divisible by 3 if and only if there exists an integer $k$ such that $n=3k+1$ or $n=3k+2$. Also using the definition $a$ divides $b$ written $a$|$b$ if $b=ac$ for some $c$ in integers.

Here it goes:

$$3\nmid n$$

$$3k+1=3n$$

$$\frac{3k+1}{3}=n$$

I assume Euclid's lemma hasn't be introduced yet. As that would make this trivial. — fleablood, Jan 20 '19 at 02:05
Yes we have not gotten to Euclid's lemma. In a month we are covering that. For now the question is in the chapter of contrapositive and contradiction. — John, Jan 20 '19 at 02:09
@rtybase no not yet. It is only the first 2 weeks of this class — John, Jan 20 '19 at 02:12
The reason I am asking is, you mentioned $n=3k+1$ or $n=3k+2$ so you are familiar with the divisibility with remainder theorem? — rtybase, Jan 20 '19 at 02:14
I agree with rtybase... Just prove on with $n=3k+1$ or $n=3k+2$. (To write this way is just using the defn of divisibility) — Macrophage, Jan 20 '19 at 02:17
More to the point. If $3\not\mid n$ then $n \ne 3k$ for any $k$. So $n = 3k + 1$ or $n = 3k + 2$. (That's the remainder theorem). So then $n^2 = ???$. Obviously $3k + 1 = 3n$ NEVER happens for integers so ... anything you prove from this is doomed to be garbage. — fleablood, Jan 20 '19 at 02:22
I am utterly perplexed as to where the $3n$ came from at all. — fleablood, Jan 20 '19 at 02:31
@fleablood "Also using the definition a divides b written a|b if b=ac for some c in integers." Is that not an applicable way to use that definition? — John, Jan 20 '19 at 02:33
Write so $3|n^2$ means there exists a $k$ so that $n^2 = 3k$. But how the #### did you conclude that there is a $k$ so that $3k + 1 = 3n$? — fleablood, Jan 20 '19 at 05:42
Would it be possible to just say $3\mid n^2\Rightarrow n^2\equiv0\pmod{3}\Rightarrow n^2-0^2\equiv0\pmod{3}\Rightarrow (n-0)(n+0)\equiv0\pmod{3}\Rightarrow$$n\equiv0\pmod{3}\Rightarrow 3\mid n$? — , Aug 22 '21 at 03:34

score 2 · Answer 1 · answered Jan 20 '19 at 02:47

You don't need to resort to proving the contrapositive; it's possible to prove the statement directly:

If we take as given that $3$ divides (exactly) one of the three consecutive numbers $n-1$, $n$, and $n+1$, then $3$ divides their product, $(n-1)n(n+1)=n^3-n$. Now if $3$ divides $n^2$, then it also divides $n^3$, and thus it divides the difference, $n^3-(n^3-n)=n$.

score 1 · Answer 2 · answered Jan 20 '19 at 02:17

Presumably you have had the division theorem.

For $n$ and integer there exist integer $k, r$ so that $n = 3k + r$ where $0 \le r < 3$. So $r = 0, 1$ or $2$.

Can you accept that?

If $r=0$ then $3|n$.

If $r = 1$ then $n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ and $3\not \mid n^2$ and that's a contradiction.

If $r = 2$ then $n^2= (3k + 2)^2 = 9k^2 + 12k + 4 = 3(k^2 + 4k+ 1) + 1$ and $3\not \mid n^2$ and that's a contradiction.

So if $3|n^2$ then the only possibility is $n = 3k$ for some $k$.

John Omielan · Answer 3 · 2019-01-20T02:15:19.120

0

I believe a simpler way is to use the unique factorization to show that for

$$n = \prod_{i \, = \, 1}^{m} p_i^{a_i} \tag{1}\label{eq1}$$

where the $p_i$ are unique primes, means that

$$n^2 = \prod_{i \, = \, 1}^{m} p_i^{2a_i} \tag{2}\label{eq2}$$

Thus, if $3 \; \vert \; n^2$, then one of the $p_i$ must be $3$, so $3 \; \vert \; n$.

Also, using your suggestion, if $n = 3k + 1$ or $n = 3k + 2$, then $n^2 = 9k^2 + 6k + 1$ or $n^2 = 9k^2 + 12k + 4$. In either case, when dividing by $3$, there is a remainder of $1$, showing that $3$ doesn't divide $n^2$. But, as $3 \; \vert \; n^2$, then $n$ cannot be either of the $2$ forms, so it must be $n = 3k$, giving that $3 \; \vert \; n$. This is an example of proving what's requested using contrapositive as it shows that if $3 \not{\vert} \; n$, then $3 \not{\vert} \; n^2$.

edited Jan 20 '19 at 02:15

answered Jan 20 '19 at 01:59

John Omielan

47,976

Thanks for you comment. We have not gotten to unique factorization yet in mathematical reasoning class. Would my proof suffice? – John Jan 20 '19 at 02:00
@John I was editing my response when you wrote your comment. I hope it sufficiently answers your question. – John Omielan Jan 20 '19 at 02:02
@John Based on the definition for contrapositive, I believe what I showed in the last paragraph uses the contrapositive technique. As for what you have in your question, as one of the comments state, I'm not sure how you get that $3k + 1 = 3n$, i.e., where does the "$3$" part come from in $3n$? – John Omielan Jan 20 '19 at 02:17

score 0 · Answer 4 · answered Jan 20 '19 at 03:00

In If $n \mid a^2 $, what is the largest $m$ for which $m \mid a$?, I prove this result:

Given $n = \prod p_i^{a_i}$, then the largest $m$ such that $m | a$ for all $a$ such that $n | a^2$ is $m = \prod p_i^{\lceil \frac{a_i}{2}\rceil}$.

If $n$ is a prime, $3$ in this problems case, then $n = 3^1$ so $m = 3^1$ also.

Prove that for all integers $n$ if $3 \mid n^2$, then $3 \mid n$

4 Answers4